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2 years ago

What plane is this? i want the model

Engineering

1 answer:

myrzilka [38]2 years ago

4 0

Answer:

I think that plane's name is

KLM ....Because you can see the mysterious Leter in the plane's

Forward and the back of the plane

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Before cutting coarse screw threads, the operator should lubricate: A. The leadscrew and gearbox B. The ways and cross slide C.

Maksim231197 [3]

Answer:

(d) all of the above

Explanation:

before cutting the screw threads the operator should lubricate all of the machine parts given in the option that is lead screw and gearbox , the ways and the cross slide and the carriage and half-nuts. we should use lubrication because it reduces the overall system friction and if friction is reduced then heat generated due to friction is also decreases which is beneficial

so option (D) will be correct because we need lubricate in all the given parts

8 0

2 years ago

PLZ ASAP WILL GIVE BRAINLIST

gavmur [86]

Answer: A, B, C & F (interacting w computers, making decisions & solving problems, evaluating information & getting information).

Explanation: Those are the correct & verified answers.

7 0

2 years ago

Read 2 more answers

A south-facing collector at latitude 40◦ is tipped up at an angle equal to its latitude. Compute the following insolations for J

BartSMP [9]

Answer:

Explanation:

(c). looking for the radiation of the collector is given thus

C = 0.095 + 0.04 sin [360/365(n-100)] = 0.095 + 0.04 sin [360/365(1-100)]

C = 0.05535

∴ Diffuse radiation of the collector Idc = C*Ib + (1+cosσ/2)

Idc = 0.5535 * 908.7 (1+cos40/ 2) = 44.41 W/m²

Idc = 44.41 W/m²

3 0

2 years ago

An inventor claims to have developed a food freezer that at steady state requires a power input of 0.25 kW to extract energy by

Serga [27]

Answer:

investor claim is acceptable

Explanation:

given data

Win = 0.25 kW

Qc = 3000 J/s = 3kW

Th = 293 K

Tc = 270 K

solution

we get here coefficient of performance of cycle is

coefficient of performance = $\frac{Qc}{Win}$ ..................1

put here value and we get

coefficient of performance = $\frac{3}{0.25}$

coefficient of performance = 1.2

and

coefficient of performance of ideal refrigeration is

coefficient of performance = $\frac{Tc}{Th-Tc}$ ..................2

coefficient of performance = $\frac{270}{293-270}$

coefficient of performance = 11.74

and

we can see here that coefficient of performance of ideal refrigeration is is more than real cycle coefficient of performance

so investor claim is acceptable

6 0

2 years ago

1. For ball bearings, determine: (a) The factor by which the catalog rating (C10) must be increased, if the life of a bearing un

ElenaW [278]

Answer:

(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.

Equation 11-1: F*L^(1/3) = Constant

Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483

Explanation:

(a)The Catalog rating(C)

Bearing life: $L_1 = L , L_2 = 2L$

Catalog rating: $C_1 = C , C_2 = ? ,$

From given equation bearing life equation,

$F\times\frac{1}{3} (L_1) = C_1 ...(1) \\\\ F\times\frac{1}{3} (L_2) =C_2...(2)$

we Dividing eqn (2) with (1)

$\frac{C_2}{C_1} =\frac{1}{3} (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\ C_2 = 1.26 C$

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

$R_1 = 0.9 , R_2 = 0.99$

Now calculating life adjustment factor for both value of reliability from Weibull parametres

$a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}$

$= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\ = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968$

Similarly

$a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\ = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\ = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215$

Now calculating bearing life for each value

$L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L$

Now using given ball bearing life equation and dividing each other similar to previous problem

$\frac{C_2}{C_1} = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\ C_2 = C* (\frac{0.2215L }{0.9968L} )^{1/3}\\\\ C_2 = 0.61 C$

Catalog rating increased by factor of 0.61

6 0

3 years ago